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<br />TRIGONOMETRIC FORMUL.'E
<br />B B B
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<br />Right Triangle Oblique Triangles
<br />Solution of Right Triangles
<br />For Angle A. sin = a ; coo= c b , tan= a , cot = b a' sec = , cosec =
<br />c b b a
<br />Given Required
<br />a,b A, B,° tanA=b=cotB,c= ar-} z=a 1 E ¢Z
<br />F-42
<br />a, a A, B, b sin A = e = cos B, b=%,1o -(-a) (c—a) = c AJ ]. — o i
<br />A, a B, b, c B=90°—A, b= a cotA, a= a
<br />sin A.
<br />A b B, a, a B=90°—A a= b tan A c= b
<br />i cos A,
<br />A, c B, a, b B= 90°—A, a = ° sin A, b = c cos A,
<br />Solution of Oblique Triangles
<br />i. Given Required a sin B in C
<br />A, B,a b, c, C b= s .C=180°—(A+B)in A,c= Sin
<br />A,a,b B,c,C sing= b Sin A ,Ca sin C
<br />a =180°—(A } B),c= sin
<br />a, b, C A, B, c A+B=180°—C, tan Ij(A—B)=(a—b)tan (A+B)
<br />a b
<br />� a sin C +
<br />C =
<br />Sin A
<br />a, b, c A, B, C s,—a+2+a,SSin ljA=.`II (9 b(c—a
<br />sing= a ), C--180--(A+-B)
<br />e
<br />a, b, c Area s= a -{-b+ 2 ,erea
<br />A, b, c Area area = b e sin A
<br />2
<br />aE sin B sin C
<br />A, B, C, a Area area = 2 sin A
<br />REDUCTION TO HORIZONTAL
<br />Horizontal distance=Slope distance multiplied by the
<br />cosine of the vertical angle. Thus: slope distance =310.4 ft.
<br />S�Qnpe Vert. angle =50 101. From Table, Page IX. cos 5' lOO'=
<br />e M. Horizontal distance=319.4X.9959=316.09 ft.
<br />i 510 PQg1e Horizontal distance also= Slope distance minus slope
<br />Ve distance times (1—cosine of vertical angle). With the
<br />same figures as in the preceding example, the follow -
<br />Horizontal distance ina result is obtained. Cosine 60101=.9959.1—.9959=.0041.
<br />319.4X.0041=1.31. 319.4-1.31=318.09 It.
<br />When the rise is known, the horizontal distance is approximately:—the slope dist-
<br />ance less the square of the rise divided by twice the slope distance. Thus: rise=l4 ft,
<br />slope distance=3026 ft Horizontal distance=3026— 14 X 14 =3026-0 32=30228 fY.
<br />2 X 302.6
<br />JMADE IN U. 8. A.
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